Evaluate : ∫ cot−1 {1 − x+ x^ 2 } x ∈ [1 0] .

Question

Evaluate : \(\int\limits_{0}^{1}\) cot−1 {1 − x + x² } x.

Answer 1

= 2(tan ^-1 1 - \(\frac{1}{2} log 2\)) - \((0 - \frac{1}{2}log 1)\) (By second fundamental theorem of integral calculus)

 = \(2(\frac{{\pi}}{4} - \frac{1}{2} log\,2) = \frac{\pi}{2} - log \,2\) ( \(\because\) tan⁻¹ 1 = \(\frac{\pi}{4}\) and log 1 = 0)

Hence, \(\int\limits_{0}^1\)cot^-1{1 - x + x²} dx = \(\frac{\pi}{2} - log 2.\)